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If $f: \mathbb{R}^n \rightarrow \mathbb{R}$ is smooth and compactly supported, one has $$\int |\Delta f(\mathbf{x})|^2\,d\mathbf{x} = \int \| Hf(\mathbf{x}) \|_F^2\,d\mathbf{x}\,,$$ where $\Delta$ denotes the Laplacian, $H$ denotes the Hessian (matrix of second derivatives), and $\| \cdot\|_F^2$ denotes the Frobenius norm squared (the sum of squares of matrix entries). What can be said about compact Lie groups?

Specifically, I have two questions about this phenomenon:

Nomenclature: In this case where all norms are $L^2$, does this equality have a generally accepted name? What about the case when other norms are introduced?

Lie groups: I would like to understand in what sense this generalizes to compact Lie groups. What can be said about smooth $f: G \rightarrow \mathbb{R}$ (or $\mathbb{C}$)?

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  • $\begingroup$ @Steven Gubkin It is discussed briefly in arxiv:0704.2833 (see the 2nd page). To see it in $\mathbb{R}^2$ you can cancel the terms of the form $(\partial^2 f/\partial x_i^2)^2$ from both sides and then judiciously apply integration by parts to each of the remaining terms: this results in a portion that cancels (as $f$ is compactly supported) and the integral of the product of a first and third derivative. $\endgroup$
    – acr
    Commented Nov 20, 2014 at 22:20

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This has nothing to do with Lie groups, I believe. Let $M$ be a Riemannian manifold. The Bochner formula on $1$-forms states that $$\nabla^* \nabla \omega = (d \delta + \delta d)\omega - \mathrm{Ric}\,\omega.$$ Hence we have for any compactly supported function $f$ (writing round brackets for the $L^2$ scalar product) $$\|Hf\|_{L^2}^2 = \|\nabla d f\|_{L^2}^2 = (\nabla^* \nabla d f, df) = ((d \delta + \delta d - \mathrm{Ric})d f, df ) = (d\delta d f, d f) - (\mathrm{Ric}\, df, df)$$ Using that on functions, $\Delta f = \delta d f$, this is equal to $$(d \Delta f, df) - (\mathrm{Ric}\, df, df) = \|\Delta f\|_{L^2}^2 - (\mathrm{Ric}\, df, df)$$ This is a standard formula of Riemannian geometry.

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  • $\begingroup$ I was not aware of this beautiful, generic expansion in terms of the Ricci tensor. So...this reduces my question to understanding the Ricci tensor over Lie groups! $\endgroup$
    – acr
    Commented Nov 21, 2014 at 11:52
  • $\begingroup$ I think the signs in front of the Ricci tensor are wrong in the second and third displayed equation; also an appearance of $f$ is missing in the third displayed equation. $\endgroup$
    – acr
    Commented Dec 8, 2014 at 15:25
  • $\begingroup$ @MatthiasLudewig: Do you know whether these formulae extend to connections in Hermitian bundles (with, obviously, the Ricci tensor replaced by something appropriate)? $\endgroup$
    – Alex M.
    Commented Jul 4, 2019 at 7:00
  • $\begingroup$ What exactly would you like a formula for in this context? $\endgroup$
    – Matthias Ludewig
    Commented Jul 9, 2019 at 7:36

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